Question

calculus help...plz explain and answer question...i totally dont know what they are...studying on my own

Answer 1

http://assets.openstudy.com/updates/attachments/517d65e4e4b0be6b54ab31bc-best_mathematician-1367172592503-3.png

Answer 2

Do you understand vectors and parametric equations?

Answer 3

i know a little abput para,etric equations

Answer 4

The position of the hoop is given by
\[\vec{r} = (t-\sin(t))\vec{i}+(1-\cos(t))\vec{j}\

So |v| will be the magnitude of the derivative of r
and |a| will be the magnitude of the derivative of v

Answer 5

ok. so how to solve for first and second answer

Answer 6

Do you know how to take derivatives?

Answer 7

yes sir...i know how to do everything except last two chapters of calculus

Answer 8

well first you're going to be taking
\[\frac{d \vec{r}}{dt}\]

Answer 9

of what

Answer 10

Of the vector r that you're given.

To take a derivative of a vector take the derivative of its components and still leave them attached to the components.

So for example the derivative of the vector s with components (3t^2 +2)i + 2j
would be ds/dt = (6t)*i + 0j

Answer 11

To take a derivative of a vector take the derivative of its components and still leave them attached to their unit vectors**.

Answer 12

so is the derivative (-costi)+(sintj)

Answer 13

the i component resolves to (1-cos(t))*i

Answer 14

i is just x and j is just y...right?

Answer 15

Yup

Answer 16

so the derivative would be - wait what is t then

Answer 17

t is telling you how x and y changes -- t is time

Answer 18

ohk...would you give me the derivative of first part (t+sint)i...i will get the rest of the part..its just confusing me

Answer 19

Well when you were taking derivatives as a function of x, you were seeing how things changed when you moved x. Now you're just seeing how they move when you change the time.

dr/dt = v = d((t+sint)i+(1-cos(t))j)/dt = (1 - cos(t))i + (sin(t))j

Answer 20

ohk got it...thx...now i know to set them as zero and then solve but again i and j is confusing me

Answer 21

Well some text books use the letters x and y instead of i and j

Answer 22

Just be sure to write an arrow over them if you do use them

Answer 23

lemme try (1-cost)i + (sint)j = 0
(1-cost)i = - (sint)j
(1-cost)/(sint) = - j / i

Answer 24

You're treating i and j like regular variables. They signify a direction (x and y). 1/i has no meaning if i is a vector.

(1-cos(t))i Just says that its velocity in the x direction is 1-cos(t) (slowing down and speeding up)

for instance the vector
\[\vec{m} = 4\vec{i} + 6\vec{j}\]
visually looks like this



Your answer for v will simply be
(1 - cos(t))i + (sin(t))j